Question

If A and B are non-singular matrices, then [MP PET 1991; Kurukshetra CEE 1998]
A. ${{(AB)}^{-1}}={{A}^{-1}}{{B}^{-1}}$
B. $AB=BA$
C. $(AB{)}'={A}'{B}'$
D. ${{(AB)}^{-1}}={{B}^{-1}}{{A}^{-1}}$

Answer 1

Hint: In this question, we have to check for the option satisfying the properties of non-singular invertible matrices. A non-singular matrix is created when two non-singular matrices are combined. If the identity matrix is the product of the matrix and its inverse, then an n-by-n square matrix is said to be invertible (also known as non-singular or non-degenerate) in linear algebra.

Complete step-by-step solution:
It is given that matrices $A$ and $B$ are non-singular so, they both have an inverse, then AB will also have an inverse,
${{(AB)}^{-1}}={{B}^{-1}}{{A}^{-1}}$

So, option D is correct.

Additional Information: Properties of Invertible Matrices
Assume that $A$ and $B$ are $n \times n$ invertible matrices. Then:
$AB$ is invertible $(AB)^{-1}=B^{-1}A^{-1}$.
$A^{-1}$ is invertible $(A^{-1})^{-1}=A$.
$kA$ is invertible for any nonzero scalar $k (kA)^{-1}=\dfrac{1}{k}A^{-1}$.
If A is a diagonal matrix, with diagonal entries $d_{1},\: d_{2},\cdots, d_{n}$ where none of the diagonal entries is 0, then $A^{−1}$ exists and is a diagonal matrix. Furthermore, the diagonal entries of $A^{−1} $ are $1/d_{1},\: 1/d_{2},\cdots , 1/d_{n}$.
If product $AB$ is not invertible, then $A$ or $B$ is not invertible.
If $A$ or $B$ are not invertible, then $AB$ is not invertible.

Note: If the identity matrix is the product of the matrix and its inverse, then an n-by-n square matrix is said to be invertible (also known as non-singular or non-degenerate) in linear algebra. One who has a non-zero determinant has a non-singular matrix. The matrix must satisfy all the conditions of the invertible matrix theorem. To determine whether the matrix is single or non-singular, this check is necessary.